Let $\{f_{n}\}$ be this sequence of functions: $f_{n}(x)=nx$ when $0\leq x\leq \frac{1}{n}$, $f_{n}(x)=2-nx$ when $\frac{1}{n}<x<\frac{2}{n}$ and 0, when $\frac{2}{n} \leq x \leq 1$.
I have to prove that it is not uniformly convergent to $f \equiv 0 $.
My attempt:
By contradiction, suppose it is uniformly convergent. Then for $\epsilon=1/2 >0, \exists n_{0} \in \mathbb{N}$ such that $\forall n\geq n_{0}, |f_{n}(x)|<1/2 \ \ \forall x\in[0,1]$ In particular for $n_{0}$ we have $|f_{n_{0}}(x)|<1/2 \ \ \forall x\in[0,1]$.
Choosing $x=\frac{1}{n_{0}} \in [0,1]$, we have$|f_{n_{0}}(x)|=1$, which is a contradiction.
Is it correct ?
That's correct! For any uniform convergence to zero problem you can just find a point of each function that is a fixed height above zero, which means their supremum norms can never converge to zero.